A seconds pendulum is suspended from the ceiling of a trolley moving horizontally with an acceleration of `4 m//s^(2)` . Its period of oscillation is

Options

A. `(a)1.90s`
B. `(b)1.70 s`
C. `(c)2.30 s`
D. `(d)1.40 s`

Explanation

To find the period of oscillation of a seconds pendulum suspended from the ceiling of a trolley moving horizontally with an acceleration of \(4 \, \text{m/s}^2\), we can follow these steps:

### Step 1: Understand the Concept of a Seconds Pendulum
A seconds pendulum is defined as a pendulum that takes exactly 2 seconds to complete one full oscillation (one complete swing back and forth). Therefore, the period \(T\) of a seconds pendulum is \(2 \, \text{s}\).

### Step 2: Use the Formula for the Period of a Pendulum
The formula for the period of a simple pendulum is given by:
\[ T = 2\pi \sqrt{\frac{L}{g}} \]
where:
- \(T\) is the period of the pendulum,
- \(L\) is the length of the pendulum,
- \(g\) is the acceleration due to gravity.

### Step 3: Calculate the Length of the Pendulum
Since we know that for a seconds pendulum \(T = 2 \, \text{s}\), we can set up the equation:
\[ 2 = 2\pi \sqrt{\frac{L}{g}} \]
Squaring both sides, we get:
\[ 4 = 4\pi^2 \frac{L}{g} \]
Rearranging gives:
\[ L = \frac{g}{\pi^2} \]
Assuming \(g \approx 10 \, \text{m/s}^2\):
\[ L = \frac{10}{\pi^2} \approx 1 \, \text{m} \]

### Step 4: Determine the Effective Acceleration
When the trolley accelerates horizontally, the effective acceleration due to gravity \(g'\) can be calculated using the Pythagorean theorem:
\[ g' = \sqrt{g^2 + a^2} \]
where \(a\) is the horizontal acceleration of the trolley. Substituting \(g = 10 \, \text{m/s}^2\) and \(a = 4 \, \text{m/s}^2\):
\[ g' = \sqrt{10^2 + 4^2} = \sqrt{100 + 16} = \sqrt{116} \approx 10.77 \, \text{m/s}^2 \]

### Step 5: Calculate the New Period of the Pendulum
Now, we can find the new period \(T'\) using the effective acceleration \(g'\):
\[ T' = 2\pi \sqrt{\frac{L}{g'}} \]
Substituting \(L \approx 1 \, \text{m}\) and \(g' \approx 10.77 \, \text{m/s}^2\):
\[ T' = 2\pi \sqrt{\frac{1}{10.77}} \approx 2\pi \cdot 0.303 \approx 1.9 \, \text{s} \]

### Final Answer
The period of oscillation of the seconds pendulum in the moving trolley is approximately \(1.9 \, \text{s}\).

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